JEE Main 31-Aug-2021 Shift 2 Solved Paper

Given, number of turns, n=1000m−1
Electric current, i = 0.75 A
Relative permeability of 1st material
µ1=500
Relative permeability of 2nd material
µ2=750
we know that µr=1+χ⇒χ
=µr−1
For 1st and 2nd material,
χ1=χ1−1=500−1=499
and χ2=µ2−1=750−1=749
Now, magnetic intensity, M = ni
Let M1 and M2 be the magnetisation and m1 and m2 be the magnetic moments of material 1st and 2nd.
Using,

I1

M

=χ1 ...(i)
and

I2

M

=χ2 ...(ii)
Dividing Eqs. (i) and (ii),

I1

I2

=

χ1

χ2

...(iii)
We also know that, i.e I=

m

V

I1

I2

=

m1

m2

...(iv)
From Eqs. (iii) and (iv), we get

m1

m2

=

χ1

χ2

=

499

749

Hence, fraction change in magnetic moment,

Δm

m1

=

m2−m1

m1

=(

m2

m1

−1)
=(

749

499

−1)=

250

499

Thus, value of χ is 250.